k - spsc - Graz University of Technology

SPSC – Signal Processing & Speech Communication Lab 

Wavelet Transform and 

its relation to 

multirate filter banks 

Christian Wallinger 

ASP Seminar 12 th June 2007 

Graz University of Technology, Austria 

Professor Georg Holzmann, Horst Cerjak, Christian 19.12.2005 Wallinger 12.06.07 Wavelet T. - Relation to Filter Banks 

1


Outline 

Short – Time Fourier – Transformation 

‣ Interpretation using Bandpass Filters 

‣ Uniform DFT Bank 

‣ Decimation 

‣ Inverse STFT and filter - bank interpretation 

‣ Basis Functions and Orthonormality 

‣ Continuous Time STFT 

Wavelet – Transformation 

‣ Passing from STFT to Wavelets 

‣ General Definition of Wavelets 

‣ Inversion and filter - bank interpretation 

‣ Orthonormal Basis 

‣ Discrete – Time Wavelet Transf. 

‣ Inverse 


2


SHORT-Time FOURIER TRANSF. 

sdfgsdfg 

yxvyxvyxcv 

fgjfghj 

figure 1: STFT processing in time 

time – frequency plot = Spectogram 

figure 2: spectogram 


3


Definition: 

X 

STFT 

( 

jω 

) ∑ ∞ e , m = 

n= 

−∞ 

x( 

n) 

v( 

n − m) 

e 

− jωn 

m . . . time shift – variable 

( typically an integer multiple of some fixed integer K) 

ω . . . frequency – variable 

− π ≤ ω < π 


4


Interpretation using Bandpass Filters 

Traditional Fourier Transform as a Filter Bank 

figure 3: Representation of FT in terms of a linear system 

e 

0 

− jω 

n 

1. Modulator : performs a frequency shift 

( 

jω 

) 

H e 

2. LTI – System : ideal lowpass filter 


5


( 

jω) 

H e 

Why is an ideal lowpass filter ? 

Impulse Response h(n) = 1 for all n 

( 

jω 

) 

− jωn 

H e h( n) 

e = 2πδ 

( ω) 

= ∑ ∞ 

−∞ 

n= 

a 

−π ≤ ω < π 

only zero - frequency passes 

every other frequency is completely supressed 

( 

jω 

) 

0 

e 

y ( n) 

= X 

for all n 


6


STFT as a Bank of Filters 

Expansion of Definiton for further insight! 

X 

STFT 

( 

jω 

) 

− jωm 

∞ e , m = e ∑ 

n= 

−∞ 

x( 

n) 

v( 

n 

− 

m) 

e 

jω( 

m−n) 

with: 

v( 

n − m) 

e 

jω( 

m−n) 

= 

v( 

−( 

m − n)) 

e 

jω( 

m−n) 

Convolution of x(n) with the impulse response of the LTI – System 

v(−n) 

e 

jωn 


7


figure 4: Representation of STFT in terms of a linear system 

In most applications, v(n) has a lowpass transform V(e jω ). 

 

jω 

v( − n) V ( e 

− ) 

0 

v( 

jω 

n 

− j( 0 

− n) 

e V ( e 

ω−ω ) ) 


8


figure 5: Demonstration of how STFT works 


9


In practice, we are interested in computing the Fourier transform at a discrete 

set of frequencies 

0 ≤ ω 0 

< ω 1 

< … < ω M-1 

< 2π 

Therefore the STFT reduces to a filter bank with M bandpass filters 

H 

k 

( e 

jω 

) = V ( e 

− j( ω−ωk 

) 

) 

figure 6: STFT viewed as a filter bank 


10


Uniform DFT bank 

If the frequencies ω k 

are uniformly spaced, then the system 

becomes the uniform DFT bank. 

The M filters are related as in the following manner 

H 

k 

z) 

H 

0 

( 

k 

zW ) 

( = 0 ≤ k ≤ M −1 

W 

= 

e 

j 2π 

− 

M 

 

H 

k 

( e 

jω 

) 

= 

H 

0 

⎛ 

⎜ 

e 

⎝ 

2π 

j( 

ω− k ) 

M 

⎞ 

⎟ 

⎠ 

H 0( e 

) = V ( e 

jω − jω 

) 

The uniform DFT bank is a device to compute the STFT at uniformely 

spaced frequencies. 


11


Decimation 

if passband width of V(e jω ) is narrow 

output signals y k 

(n) are narrowband lowpass signals 

this means, that yk(n) varies slowly with time 

According to this variying nature, one can exploit that to decimate the 

output. 

Decimation Ratio of M = moving the window v(k) by M samples at a time 

if filters have equal bandwidth 

 

n k 

= M 

maximally decimated analyses bank 

figure 7: Analysis bank with decimators 


12


Time – Frequency Grid 

Uniform sampling of both, ‘time’ n and ‘frequency’ ω 

figure 8: time – frequency grid 

Time spacing M corresponds to moving the window M units ( = samples ) at a time. 

2π 

frequency spacing of adjacent filters = M 


13


X 

Inversion of the STFT 

From traditional Fourier – viewpoint 

( 

jω 

e m) 

STFT 

, 

is the FT. from the time domain product 

x( n) 

v( 

n − m) 

 

2π 

1 

− = ∫ ( 

jω 

) 

jωn 

x( n) 

v( 

n m) 

X 

STFT 

e , m e dω 

2π 

0 


14


Another inversion formula is given by: 

2π 

∞ 

1 ⎛ 

∫ ∑ ( 

jω 

) 

* ⎞ 

= ⎜ 

( − ) 

jωn 

x( n) 

X 

STFT 

e , m v n m ⎟e 

dω 

2π 

⎝ m=−∞ 

⎠ 

0 

2 

∑ v = 1 

which is provided by ( m) 

m 

if 

∑ 

m 

( m) 

2 

v ≠ 1 

but finite divide right side of the formula by 

∑ m 

v 

( m) 

2 

but if window energy is infinite one cannot apply this formulation 


15


Filter Bank Interpretation of the Inverse 

F k 

(z) 

With as synthesis - filter 

Reconstruction can be done by the following synthesis bank: 

figure 9: synthesis – bank used to reconstruct x(n) 

typically n k 

= M for all k 


16


( n) 

The z – Transformation of is given by 

Xˆ 

M 

∑ − 1 

k = 0 

xˆ 

nk 

( z) = X ( z ) F ( z) 

k 

k 

xˆ 

M − 1 

in time – domain 

∞ 

( n) = x ( m) f ( n − n m) 

∑∑ 

k= 

0 m=−∞ 

k 

k 

k 

= 

M − 1 

∞ 

∑∑ 

k= 

0 m=−∞ 

y 

k 

jω ( n m) 

( n m) e k k 

f ( n − n m) 

k 

k 

k 

y 

k 

( n m) K STFT − Coefficien ts 

k 

Reconstruction is stable, if the filters F k 

(z) are stable! 

Perfect reconstruction will be obtained, if x ˆ( 

n) = x( n) 


17


Basis Functions and Orthonormality 

η 

km 

Functions of interest 

( n) =ˆ f ( n − n m) Kbasis 

functions 

k 

k 

For these double indexed functions ( basis functions n ), 

the orthonormality property means that 

{ ( )} 

η km 

∑ ∞ 

n= 

−∞ 

( n − n m ) f ( n − n m ) = δ ( k − k ) ( m − m ) 

* 

f 

k1 k1 

1 k 2 k 2 2 1 2 

δ 

1 2 

should be zero, except for those cases where k 

1 

= k2 

and m1 

= m2 


18


The Continuous - Time Case 

Main points: 

X 

STFT 

∞ 

∫ 

−∞ 

− jΩ 

t 

( jΩ,τ 

) = x( t) v( t −τ 

) e dt ( STFT ) 

x 

∞ 

1 

2π 

∫ 

j t 

()( t v t −τ 

) = X ( jΩ, 

τ ) e 

Ω dΩ 

( inv STFT ) 

−∞ 

STFT 

. 

x 

1 ∞ ∞ 

* 

STFT 

. 

jΩ 

t 

() t = ⎜ X ( jΩ, 

τ ) v ( t −τ 

) dτ 

⎟e 

dΩ 

( inv STFT ) 

2π 

∫⎜ 

∫ 

−∞ 

⎛ 

⎝ 

−∞ 

⎞ 

⎟ 

⎠ 


19


Choice of “Best Window” 

R oot 

M ean 

S quare 

duration of window function v(t) in 

time domain D t 

∞ 

2 1 2 2 

D t ∫ t v 

E 

−∞ 

() t 

= dt 

frequency domain D f 

∞ 

2 1 2 

2 

D f ∫ Ω V ( jΩ) 

2πE 

−∞ 

= dΩ 

with: 

2 

E . . . window energy E = v () t dt 

∫ 

Uncertainty principle: 

D t D f 

≥ 0.5 

Iff Gaussian – window, this inequality becomes an equality ! 


20


Filter Bank Interpretation 

figure 10: continuous – STFT 


21


THE WAVELET TRANSFORM 

Disadvantage of STFT 

uniform time – frequency box ( D = const., D const. ) 

t f 

= 

The accuracy of the estimate of the Fourier transform 

is poor at low frequencies, and improves as the frequency increases. 

Expected properties for a new function: 

‣ window width should adjust itself with ‘frequency’ 

‣ as the window gets wider in time, also the step sizes for 

moving the window should become wider. 

These goals are nicely accomplished by the wavelet transform. 


22


Passing from STFT to Wavelets 

Step 1: 

Giving up the STFT modulation scheme and obtain filters 

h 

k 

−k 

() t a ( ) 2 −k 

= h a t a > 1Kscaling 

factor, 

k = int eger 

in the frequency domain: 

k 

k 

( jΩ) = a H ( ja Ω) 

2 

H 

k 

all reponses are obtained by frequency – scaling of a prototype response H( 

jΩ) 


23


Example: 

H 

( jΩ) 

Assuming is a bandpass with cutoff frequencies α and β. 

Also a = 2 , β = 2α and the center frequency should be the geometrical 

mean of the two cutoff edges 

Ω 

k 

= 

2 

−k 

αβ = α 2 

−k 

2 

figure 11: frequency – response obtained by scaling process 


24


Ratio: 

bandwidth 

center − frequency 

Ω 

k 

= 

2 

−k 

2 

( β − α ) 

−k 

αβ 

= 

1 

2 

is independent of integer k 

In electrical filter theory such a system is often said to be a ‘constant Q’ system! 

( Q ... Quality factor 

center − frequency 

Q = ) 

bandwidth 


25


filter ouputs can be obtained by: 

a 

−k 

2 

e 

− jΩ 

k 

∞ 

τ 

∫ 

−∞ 

x 

−k 

() t h a ( τ − t) 

( ) 

dt 

Step 2: 

k 

↑ 

→ 

bandwidth 

of 

H 

k 

( jΩ) ↓ → Samplerate ↓ 

or in time domain 

k 

↑ 

→ 

window length 

↑ 

→ 

step 

size 

↑ 


26


Therefore: 

τ = 

na 

k 

T 

nK int eger, 

a 

k 

T Kstep 

size 

hence: 

h 

( 

−k 

( 

k 

) ( 

−k 

a na T − t = h nT − a t) 

Summarizing, we are computing: 

 

X 

X 

DWT 

DWT 

, 

∞ 

∫ 

−∞ 

−k 

2 

−k 

( k, 

n) a x() t h( nT − a t) 

= dt 

∞ 

∫ 

−∞ 

k 

( k n) x( t) h ( na T − t) 

= dt 

k 

DWT...Discrete Wavelet Transform 

figure 12: Analysis bank of DWT 


27


Time Frequency Grid 

figure 13: time – frequency grid 

D t 

D f 

= 

const. 


28


General Definition of the Wavelet Transform 

∞ 

1 ⎛ t − q ⎞ 

X CWT 

( p, 

q) = ∫ x() 

t f ⎜ ⎟dt 

p −∞ ⎝ p ⎠ 

p,q ... real – valued continuous variables 

According to former definition: 

p = 

k 

a 

q 

k 

= a Tn f ( t) = h( − t) 

X 

CWT 

( p, 

q) and X ( k, 

n) KKK 

wavelet coefficients 

DWT 


29


Inversion of Wavelet Transform 

x 

( t) X ( k, 

n) ψ ( t) 

= ∑∑ 

ψ k n 

k 

( t) 

n 

DWT 

k n 

where are the basis functions 

Filter Bank Interpretation of Inversion 

Reconstruction of x(t) as a designing problem of the following synthesis filter bank 

( k, 

n) K sequence 

( jΩ) K continuous intime 

X DWT 

F k 

output of synthesis filter bank : 

k 

( t) = ∑∑X 

( k n) f ( t − a nT ) 

x ˆ 

, 

k n 

DWT 

k 

figure 14: synthesis bank 


30


All synthesis filters are again generated from a fixed prototype synthesis filter 

f(t) ( mother wavelet ) 

f 

k 

−k 

2 −k 

() t = a f ( a t) 

Substituting this in the preceding equation and assuming perfect reconstruction, we get 

with: 

−k 

() = ∑∑ ( ) ( − k 

t X k, 

n a f a t − nT ) 

2 

x 

DWT 

k n 

−k 

k 

k 

− 

2 − 

−k 

k 

() t = f () t → ψ () t = a ψ ( a t − nT ) = a ψ [ a ( t − na T )] Kset 

of basis functions 

2 

ψ 

k n 

using this, we can express each basis function in terms of the filter f k 

( t) 

ψ 

k n 

( ) ( 

k 

t = f t − na T ) 

k 


31


Orthonormal Basis 

{ ()} 

Of particular interest is the case where ψ k 

t is a set of orthonormal 

n 

functions 

Therefore, we expect: 

∞ 

∫ψ * 

k 

−∞ 

n 

() t ψ () t dt = δ ( k − l) δ ( n − m) 

l m 

using Parseval’s theorem, this becomes 

1 

2π 

∞ 

∫ 

−∞ 

Ψ 

* 

k n 

( jΩ) Ψ ( jΩ) dΩ = δ ( k − l) δ ( n − m) 

l m 

and get : 

X 

DWT 

∞ 

∫ 

−∞ 

* 

( k, n) x() t ψ () t 

= dt 

k n 


32


Comparing these results, we can conclude: 

ψ 

k n 

And in particular for k = 0 and n = 0: 

( ) ( 

k 

t = h 

* a nT − t) 

* 

ψ ( t) = ( t) = h ( − t) 

for the orthonormal case fk 

( t) = h 

* 

k ( − t) 

00 

ψ 

k 

H 

Discrete – Time Wavelet Transform 

Starting with the frequency domain relation and a scaling factor a = 2 

k 

k 

j 

( e ) ( ) ω j2 

ω 

= H e K k is a nonnegative int eger 

H 

jω 

( e ) 

for highpass and k = 1, k = 2 

figure 15: Magnitude responses 


33


Let G(z) be a lowpass with response 

Using QMF – banks 

figure 16: Magnitude – response of G(z) 

or 

its equivalent 

figure 17: 3 level binary tree-structured QMF 

figure 18: equivalent 4-channel system 


34


2 

2 4 

Responses of the filters H ( z) , G( z) H ( z ), 

G( z) G( z ) H ( z ),...... 

figure 19: combinations of H(z) and G(z) 

Defining the Discrete –Time Wavelet Transform 

M −1 

k + 

( n) = x( m) h ( 2 1 n − m) 

∑ ∞ 

m= 

−∞ 

y , 0 ≤ k ≤ M − 2 

k 

k 

M − 

( n) = x( m) h ( 2 1 n − m) ( D T WT ) 

∑ ∞ 

m= 

−∞ 

y , 

M −1 

iscrete 

ime 


35


Inverse Transform 

F 

2 

( z) = H ( z) , F ( z) H ( z ) G ( ), 

K 

0 s 1 

= 

s s 

z 

figure 20: synthesis filters 


36


For perfect reconstruction x ˆ( n) = x( n) 

we can express 

X 

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 

2 

4 

2 

2 

z = F z Y z + F z Y z + + F z Y z F z Y z 

0 0 

1 1 

... 

M −1 

M − 2 M −2 

+ 

M −1 

M −1 

M −1 

and in time domain: 

x 

M −2 

∞ 

∞ 

2k 

+ 1 

M −1 

( n) = y ( m) f ( n − 2 m) + y ( m) f ( n − 2 m) 

∑∑ 

∑ 

k k 

k= 

0 m=−∞ 

m=−∞ 

M −1 

M −1 


37


Main References 

Multirate Systems and Filter Banks 

(Prentice Hall Signal Processing Series) 

by P. P. Vaidyanathan 


38


Thank you for attention ! 


39

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